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Precision when dividing two functions going against zero

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Hi, I have two functions A=sinc(c1*x)-cos(c2*x) whichs is Zero for x=0. Now I want to calculate A/(c3*x). The expected result is exactly 1 at x=0. But numerically I get 1.05 Is there a way to increase precision in matlab? Format Long didn't work. I want to plot the behaviour of the function for a variety of x values, from large to very small
Michael Haderlein
Michael Haderlein on 28 Jul 2014
Hm, how did you get 1.05? I'd do something like this (since I don't know your c1,2,3, I set all to 1):
semilogx(x,(sinc(x)-cos(x) )./x)
This way, it converges towards 0 (for x-> 0).
David Young
David Young on 28 Jul 2014
MATLAB computes in double precision by default. It is most unlikely that your problem is anything to do with the precision of the computation. If you show the code you used to get the value of 1.05 it may be possible to explain what is happening.

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Answers (2)

Matt J
Matt J on 28 Jul 2014
Edited: Matt J on 28 Jul 2014
Use a Taylor expansion of A(x) around x=0 to evaluate A(x)/(c3*x) at x in the near neighborhood of zero.

Patrik Ek
Patrik Ek on 28 Jul 2014
Edited: Patrik Ek on 28 Jul 2014
I think that you should try to do the calculations manually. I did the calculations manually and got that the value should converge towards 0, using a 3rd order taylor expansion. When manually evaluating:
f(x) = ( sinc(c1*x)-cos(c2*x) ) / (c3*x)
the expression I got was:
f(x) = x / ( c3*factorial(3) ) * (3*c2^2-c1^2)
This will converge towards 0 which the expression did for me when x was small enough. Notice that
sinc(c1*x) = sin(c1*x) / (c1*x)
If you plot sinc(x), cos(x), 1-x, 1+x in the same plot (using hold on) around 0 ( (-0.2, 0.2) is enough I think), it becomes quite clear that it should converge towards something small. It can be seen that cos(x) ans sinc(x) is similar at that point, but not the lines.
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Matt J
Matt J on 28 Jul 2014
Edited: Matt J on 28 Jul 2014
L'hopital's rule also confirms that the value should go to zero at x=0.

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